Venn and Tree Diagrams (AA SL)
Once a question involves two events at once, you need a picture to keep track of the overlap - a Venn diagram for "and"/"or" questions, a tree diagram for sequences of draws. This page covers the addition rule and multiplying along branches, with worked examples covering both. It's part of the broader Probability topic.
12 questions on this sub-topic.
The two rules
Covered under IB syllabus reference SL4.6, which sets out how to combine events using diagrams and tables of outcomes. Neither rule below is complicated on its own - the skill is picking which diagram fits the question.
Addition rule (Venn)
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)
Subtracting the overlap stops it being double-counted. If \(A\) and \(B\) are mutually exclusive, \(P(A\cap B)=0\) and this becomes a simple addition. Not in the formula booklet - prior knowledge.
Multiplication along a tree
\(P(A\cap B)=P(A)\times P(B|A)\)
Multiply the probabilities along a single branch to find that path's probability, then add the probabilities of every branch that leads to the outcome you want. Not in the formula booklet - prior knowledge.
Need the conditional-probability formula behind \(P(B|A)\), or the full syllabus wording? See Conditional Probability or the full Probability topic page.
Worked examples
Events \(A\) and \(B\) are mutually exclusive with \(P(A)=0.3,\ P(B)=0.25.\) Find \(P(A\cup B).\)
Worked solution
Mutually exclusive events cannot both occur, so \(P(A\cap B) = 0.\) M1
\(P(A\cup B) = 0.3 + 0.25\) A1
\(= 0.55.\) A1
A bag has 5 red and 3 blue counters. Two are drawn without replacement.
(a) Find the probability both are red.
(b) Find the probability they are different colours.
Worked solution
(a) Without replacement, the second branch's probability changes: \(P(RR)=\dfrac{5}{8}\times\dfrac{4}{7}=\dfrac{20}{56}.\) M1
Multiply along the branch and simplify: \(P(RR)=\dfrac{5}{14}.\) A1
(b) Two routes give different colours: identify \(RB\) and \(BR\). M1
\(P(RB)=\dfrac{5}{8}\times\dfrac{3}{7}=\dfrac{15}{56}.\) A1
\(P(BR)=\dfrac{3}{8}\times\dfrac{5}{7}=\dfrac{15}{56}.\) A1
Sum both routes and simplify: \(P(RB)+P(BR)=\dfrac{30}{56}=\dfrac{15}{28}.\) A1
Common mistakes
- Using the addition rule without subtracting the overlap. \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\) - dropping the last term double-counts every outcome that's in both events, unless the events are genuinely mutually exclusive.
- Keeping the same branch probability on a second draw without replacement. Removing an item changes both the total and the count remaining of that type, so the second branch of a tree needs a new denominator, not the same fraction copied down.
- Only finding one route to an outcome that has several. "Different colours" or "at least one" often has more than one path through the tree - every valid branch has to be found and added, not just the first one you spot.
Ready to practise properly?
11 Venn-and-tree-diagram questions, marked instantly like the real exam.
Quick answers
How do you find P(A or B) using a Venn diagram?
\(P(A\cup B) = P(A) + P(B) - P(A\cap B)\). Subtracting the overlap stops it being counted twice; if \(A\) and \(B\) are mutually exclusive, \(P(A\cap B)=0\) and the formula reduces to a simple addition.
How do you calculate probabilities on a tree diagram?
Multiply the probabilities along a single branch to find that path's probability, then add up the probabilities of every branch that leads to the outcome you want. See the GDC guidance on the Probability topic page for evaluating the final expressions.